A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is $^{'} a^{'}$ , the closest approach between two atoms in metallic crystal will be :
(JEE Main 2017)

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Solution

In FCC, one of the face is like as shown in figure:
By pythagoras theorem in $Δ A B C$ ,
$2 a^{2} = 16 r^{2}$
$\Rightarrow r^{2} = \frac{1}{8} a^{2}$
$\Rightarrow r = \frac{1}{2 \sqrt{2}} a$
Distance of closest approach, $\Rightarrow 2 r = \frac{a}{\sqrt{2}}$

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A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is ′a′, the closest approach between two atoms in metallic crystal will be : (JEE Main 2017)

In FCC, one of the face is like as shown in figure: By pythagoras theorem in ΔABC, 2a2=16r2 ⇒r2=18a2 ⇒r=12√2a Distance of closest approach, ⇒2r=a√2

A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is $^{'} a^{'}$ , the closest approach between two atoms in metallic crystal will be :
(JEE Main 2017)

In FCC, one of the face is like as shown in figure:
By pythagoras theorem in $Δ A B C$ ,
$2 a^{2} = 16 r^{2}$
$\Rightarrow r^{2} = \frac{1}{8} a^{2}$
$\Rightarrow r = \frac{1}{2 \sqrt{2}} a$
Distance of closest approach, $\Rightarrow 2 r = \frac{a}{\sqrt{2}}$